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Math Help - Contingent factors method

  1. #1
    Junior Member Revy's Avatar
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    Partial Fractions decomposition

    I have an example of Partial Fractions decomposition, but cant understand it.

    Here we have \frac{2x^3+4x^2+x+2}{(x-1)^2(x^2+x+1)}
    \frac{2x^3+4x^2+x+2}{(x-1)^2(x^2+x+1)}=\frac{A}{(x-2)^2}+\frac{B}{x-1}+\frac{C_x+D}{x^2+x+1}
    \frac{2x^3+4x^2+x+2}{(x-1)^2(x^2+x+1)}=\frac{A(x^2+x+1)+B(x-1)(x^2+x+1)+(C_x+D)(x-1)^2}{(x-1)^2(x^2+x+1)}

    2x^3+4x^2+x+2=A(x^2+x+1)+B(x-1)(x^2+x+1)+(C_x+D)(x-1)^2

    And whats goes next, I totally don't get
    x^3 B+C=2
    x^2 A-2C+D=4
    x^1 A+C-2D=1
    x^0 A-B+D=2

    A=3; B=2; C=0; D=1

    So could anyone explain me that most imprtant part?

    Now that I think more, it seems that I posted it on wrong section. It's just that later that fromula will be integrated, so it should be known in calculus.
    Seems I used wrong name for this method
    Last edited by Revy; September 10th 2010 at 02:07 PM. Reason: fixed mistake
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  2. #2
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by Revy View Post
    I have an example of Partial Fractions decomposition, but cant understand it.

    Here we have \frac{2x^3+4x^2+x+2}{(x-1)^2(x^2+x+1)}
    \frac{2x^3+4x^2+x+2}{(x-1)^2(x^2+x+1)}=\frac{A}{(x-2)^2}+\frac{B}{x-1}+\frac{C_x+D}{x^2+x+1}
    \frac{2x^3+4x^2+x+2}{(x-1)^2(x^2+x+1)}=\frac{A(x^2+x+1)+B(x-1)(x^2+x+1)+(C_x+D)(x-1)^2}{(x-1)^2(x^2+x+1)}

    2x^3+4x^2+x+2=A(x^2+x+1)+B(x-1)(x^2+x+1)+(C_x+D)(x-1)^2

    And whats goes next, I totally don't get
    x^3 A+B=2
    x^2 A-2C+D=4
    x^1 A+C-2D=1
    x^0 A-B+D=2

    A=3; B=2; C=0; D=1

    So could anyone explain me that most imprtant part?

    Now that I think more, it seems that I posted it on wrong section. It's just that later that fromula will be integrated, so it should be known in calculus.
    Seems I used wrong name for this method
    well first of all that isn't true ...
    you have the line ( I didn't check all your work up there) that says :

     A+B = 2

    and you have there results  A= 3 and  B=2

    if you are to put that solutions in that equation you get :

     3+2 = 2

    which isn't true



    when done correctly you just put those A,B,C,D,E .... coefficient back to the equation and you will get partial fractions

    note that you can always check your work, if you are just that what did you get (partial fractions) solve :

     \displaystyle \frac {A}{something} \pm \frac {B}{something \; else } \pm ....

    and you must get starting fraction ... if that's not the case than you made some errors while working
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  3. #3
    Junior Member Revy's Avatar
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    Quote Originally Posted by yeKciM View Post
    if that's not the case than you made some errors while working
    Yeah =D you're right, I made mistake (thats why I do hate those undersized laptops)
    Sorry, A+B=2 should be B+C=2
    now it's fixed on original post

    Seems I made this thread uselessly, because after finding out english name of method I found some easy examples (not text-book ones, where you look and see ****). So, finally, at second day of brainstorming I learned something new =D
    Last edited by Revy; September 10th 2010 at 02:31 PM.
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  4. #4
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by Revy View Post
    Yeah =D you're right, I made mistake (thats why I do hate those undersized laptops)
    Sorry, A+B=2 should be B+C=2
    now it's fixed on original post
    now it's okay

     \displaystyle \frac {2x^3+4x^2+x+2}{(x-1)^2 (x^2+x+1)} =3 \cdot  \frac {1}{(x-1)^2} +2\cdot \frac {1}{x-1} + \frac {1}{x^2+x+1}

    Quote Originally Posted by Revy View Post
    And whats goes next, I totally don't get
    x^3 B+C=2
    x^2 A-2C+D=4
    x^1 A+C-2D=1
    x^0 A-B+D=2

    A=3; B=2; C=0; D=1

    So could anyone explain me that most imprtant part?
    now what you did there is just group those x-es with same power and look at the starting fraction and see is there any with  x^3 of there is okay, and which coefficient is with him ? okay it's 2 that means those must be equal to 2 that's way you have B+C=2 and so on ... than just solve system of equations and your off
    Last edited by yeKciM; September 10th 2010 at 02:51 PM.
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  5. #5
    Junior Member Revy's Avatar
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    Ah.. I wanted to do expanding and then collecting on paper, but now I see that it's easy to do linear system just by looking at equation.

    So that's what those x^n, 0<=x<=3 meant
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